If `f (x) = cot ^(-1)((3x -x ^(3))/( 1- 3x ^(2)))and g (x) = cos ^(-1) ((1-x ^(2))/(1+x^(2)))` then `lim _(xtoa)(f(x) - f(a))/( g(x) -g (a)), 0 ltalt 1/2` is :

Options

A. 1) `(3)/(2 (1+a^(2)))`
B. 2) `3/2`
C. 3)`(-3)/(2 (1+a^(2)))`
D. 4)`-3/2`

Answer

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Define the functions
We have:
- \( f(x) = \cot^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \)
- \( g(x) = \cos^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) \)

### Step 2: Simplify \( f(x) \)
Using the identity for the cotangent inverse, we can rewrite \( f(x) \):
- Let \( x = \tan(\theta) \), then:
\[ f(x) = \cot^{-1}\left(\frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)}\right) \]
This simplifies to:
\[ f(x) = \cot^{-1}(\tan(3\theta)) = \frac{\pi}{2} - 3\theta = \frac{\pi}{2} - 3\tan^{-1}(x) \]

### Step 3: Simplify \( g(x) \)
Similarly, for \( g(x) \):
- Using the identity for cosine inverse:
\[ g(x) = \cos^{-1}\left(\frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}\right) = \cos^{-1}(\cos(2\theta)) = 2\theta = 2\tan^{-1}(x) \]

### Step 4: Find the derivatives
Now we need to find the derivatives \( f'(x) \) and \( g'(x) \):
- From \( f(x) = \frac{\pi}{2} - 3\tan^{-1}(x) \):
\[ f'(x) = -3 \cdot \frac{1}{1 + x^2} \]
- From \( g(x) = 2\tan^{-1}(x) \):
\[ g'(x) = 2 \cdot \frac{1}{1 + x^2} \]

### Step 5: Evaluate the limit
We need to evaluate:
\[ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = \frac{f'(a)}{g'(a)} \]
Substituting the derivatives:
\[ \frac{f'(a)}{g'(a)} = \frac{-3/(1 + a^2)}{2/(1 + a^2)} = \frac{-3}{2} \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} = -\frac{3}{2} \]

### Final Answer
The correct option is:
- **Option 4: \(-\frac{3}{2}\)**